标签：leetcode
Given an array of n integers where n > 1, nums,
return an array output such that output[i] is
equal to the product of all the elements of nums except nums[i].
Solve it without division and in O(n).
For example, given [1,2,3,4],
return [24,12,8,6].
Follow up:
Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)
sdf
前几天刚看facebook的面经出现这题， leetocde就加上了， 不能用除法， 则维护当前元素左边所有元素的乘积以及右边所有元素的乘积， 相乘得到 product of array except self !
Because we cannot use division, so assume we have two integer arrays with the same length of nums, int[]
leftProd = new int[nums.length]; int[] rightProd = new int[nums.length], we store the product of all the left elements in leftProd and
the product of all the right elements in rightProd,
then the product of leftProd[i] and rightProd[i] will
be the value we want to put into the result. take the example of num[] = {2, 4, 3, 6},
thenleftProd will
be {1, 2, 8, 24} ,
and rightProd will be {72, 18, 6, 1}.
public class Solution {
public int[] productExceptSelf(int[] nums) {
if(nums == null)
return null;
int[] res = new int[nums.length];
for(int i = 0; i < nums.length; i++){
if(i == 0)
res[i] = 1;
else
res[i] = res[i - 1] * nums[i - 1];
}
int prod = 1;
for(int i = nums.length - 1; i >= 0; i--){
res[i] = res[i] * prod;
prod *= nums[i];
}
return res;
}
}
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#leetcode#Product of Array Except Self
标签：leetcode
原文地址：http://blog.csdn.net/chibaoneliuliuni/article/details/46908181
